Question

What is the mass of precipitate formed when 50 ml of 16.9% w/v solution of AgNO3 is mixed with 50 ml of 5.8% w/v NaCl solution?

Answer 1

Well, `"w/v"-="Mass of solute"/"Volume of solution"`, we should get over `7*g` of silver chloride.

Explanation:

And thus `"mass of silver nitrate"=16.9%xx50*mL=8.45*g`.

And this represents a molar quantity of `(8.45*g)/(169.87*g*mol^-1)`

`=0.0497*mol`, with respect to `AgNO_3`.

Likewise `"mass of sodium chloride"=5.8%xx50*mL=2.90*g`.

And this represents a molar quantity of `(2.90*g)/(58.44*g*mol^-1)`

`=0.0497*mol` with respect to `NaCl`.

Clearly the reagents are present in 1:1 molar ratio. The reaction that occurs in solution is the precipitation of a curdy white mass of `AgCl(s)`, i.e. and we represent this reaction by the net ionic equation.....

`Ag^(+) + Cl^(-)rarrAgCl(s)darr`

Of course the complete reaction is....

`AgNO_3(aq) + NaCl(aq) rarr AgCl(s)darr + NaNO_3(aq)`, i.e. sodium nitrate remains in solution and can be separated (with effort) from the precipitate.

And given the stoichiometry, we gets `0.04974*molxx143.32*g*mol^-1=7.11*g`.

Of course a material such as silver halide would be very hard to isolate. Particle size is very small; it is likely to clog the filter, and filter very slowly; and moreover `AgCl` is photoactive, and would decompose under light.